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Gravitation question

2019 · 12 Jan · Shift 1 · Q49
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Gravitation question

2019 · 12 Jan · Shift 1 · Q49

JEE MainPhysicsGravitationMCQ+4 / −1
A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be :
  1. A
    in the same circular orbit of radius R
  2. B
    such that it escapes to infinity
  3. C
    in a circular orbit of a different radius
  4. D
    in an elliptical orbit
View written solutionFree

Correct answer: D

  1. Initial speed of the satellite

For a circular orbit of radius RRR around Earth,

vs=GMERv_s = \sqrt{\frac{GM_E}{R}}vs​=RGME​​​

where GGG is the gravitational constant and MEM_EME​ is the mass of the Earth.

The problem states that the meteorite has the same speed just before collision:

vm=vs=GMERv_m = v_s = \sqrt{\frac{GM_E}{R}}vm​=vs​=RGME​​​


  1. Direction of velocities just before collision
  • The satellite in circular orbit moves tangentially to the orbit.
  • The meteorite is falling towards Earth, so its velocity is radially inward.

Thus, the two velocities are perpendicular.


  1. Velocity after completely inelastic collision

Both bodies have mass MMM, and they stick together. By conservation of linear momentum:

V⃗=Mv⃗s+Mv⃗m2M=v⃗s+v⃗m2\vec V = \frac{M\vec v_s + M\vec v_m}{2M} = \frac{\vec v_s + \vec v_m}{2}V=2MMvs​+Mvm​​=2vs​+vm​​

Since v⃗s⊥v⃗m\vec v_s \perp \vec v_mvs​⊥vm​ and ∣v⃗s∣=∣v⃗m∣=v|\vec v_s|=|\vec v_m|=v∣vs​∣=∣vm​∣=v,

∣V⃗∣=12v2+v2=v2|\vec V| = \frac{1}{2}\sqrt{v^2+v^2} = \frac{v}{\sqrt{2}}∣V∣=21​v2+v2​=2​v​

So after collision, the combined body has speed

V=12GMER=GME2RV = \frac{1}{\sqrt{2}}\sqrt{\frac{GM_E}{R}} = \sqrt{\frac{GM_E}{2R}}V=2​1​RGME​​​=2RGME​​​


  1. Can the new orbit be circular?

For a circular orbit at radius RRR, the required speed is

vc=GMERv_c = \sqrt{\frac{GM_E}{R}}vc​=RGME​​​

But the new speed is

V=GME2R<GMERV = \sqrt{\frac{GM_E}{2R}} < \sqrt{\frac{GM_E}{R}}V=2RGME​​​<RGME​​​

So it cannot remain in the same circular orbit, nor can it move in another circular orbit at that instant because the velocity also has a radial component.

Hence, options A and C are ruled out.


  1. Can it escape to infinity?

Escape speed at radius RRR is

vesc=2GMERv_{esc} = \sqrt{\frac{2GM_E}{R}}vesc​=R2GME​​​

Clearly,

V=GME2R<2GMER=vescV = \sqrt{\frac{GM_E}{2R}} < \sqrt{\frac{2GM_E}{R}} = v_{esc}V=2RGME​​​<R2GME​​​=vesc​

So it cannot escape.

Thus, option B is ruled out.


  1. Nature of the orbit from energy

Specific mechanical energy after collision is

ϵ=V22−GMER\epsilon = \frac{V^2}{2} - \frac{GM_E}{R}ϵ=2V2​−RGME​​

Using V2=GME2RV^2 = \frac{GM_E}{2R}V2=2RGME​​,

ϵ=12⋅GME2R−GMER=GME4R−GMER=−3GME4R<0\epsilon = \frac{1}{2}\cdot \frac{GM_E}{2R} - \frac{GM_E}{R} = \frac{GM_E}{4R} - \frac{GM_E}{R} = -\frac{3GM_E}{4R} < 0ϵ=21​⋅2RGME​​−RGME​​=4RGME​​−RGME​​=−4R3GME​​<0

Negative total energy means the body remains in a bound orbit.

Since it is not circular and is bound, the orbit must be elliptical.


  1. Checking options
  • A: Same circular orbit of radius RRR — False
  • B: Escapes to infinity — False
  • C: Circular orbit of different radius — False
  • D: Elliptical orbit — True

Final Answer

D: in an elliptical orbit\boxed{\text{D: in an elliptical orbit}}D: in an elliptical orbit​

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